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# Check whether the binary equivalent of a number ends with given string or not

• Last Updated : 16 Jul, 2021

Given a positive integer N, the task is to check whether the binary equivalent of that integer ends with the given string str or not.
Print “Yes” if it ends in “str”. Otherwise, Print “No”.
Examples:

Input: N = 23, str = “111”
Output: Yes
Explanation:
Binary of 23 = 10111, which ends with “111”

Input: N = 5, str = “111”
Output: No

Approach: The idea is to find the Binary Equivalent of N and check if str is a Suffix of its Binary Equivalent.

Below is the implementation of the above approach:

## C++

 // C++ implementation of the// above approach #include using namespace std; // Function returns true if// s1 is suffix of s2bool isSuffix(string s1, string s2){    int n1 = s1.length();    int n2 = s2.length();    if (n1 > n2)        return false;    for (int i = 0; i < n1; i++)        if (s1[n1 - i - 1] != s2[n2 - i - 1])            return false;    return true;} // Function to check if binary equivalent// of a number ends in "111" or notbool CheckBinaryEquivalent(int N, string str){     // To store the binary    // number    int B_Number = 0;    int cnt = 0;     while (N != 0) {         int rem = N % 2;        int c = pow(10, cnt);        B_Number += rem * c;        N /= 2;         // Count used to store        // exponent value        cnt++;    }     string bin = to_string(B_Number);    return isSuffix(str, bin);} // Driver codeint main(){     int N = 23;    string str = "111";    if (CheckBinaryEquivalent(N, str))        cout << "Yes";    else        cout << "No";     return 0;}

## Java

 // Java implementation of the// above approachclass GFG{ // Function returns true if// s1 is suffix of s2static boolean isSuffix(String s1, String s2){    int n1 = s1.length(),        n2 = s2.length();             if (n1 > n2)        return false;             for(int i = 0; i < n1; i++)        if (s1.charAt(n1 - i - 1) !=            s2.charAt(n2 - i - 1))        return false;             return true;} // Function to check if binary equivalent// of a number ends in "111" or notstatic boolean CheckBinaryEquivalent(int N,                                     String str){     // To store the binary    // number    int B_Number = 0;    int cnt = 0;     while (N != 0)    {        int rem = N % 2;        int c = (int)Math.pow(10, cnt);        B_Number += rem * c;        N /= 2;         // Count used to store        // exponent value        cnt++;    }         String bin = Integer.toString(B_Number);    return isSuffix(str, bin);} // Driver codepublic static void main(String[] args){    int N = 23;    String str = "111";         if (CheckBinaryEquivalent(N, str))        System.out.print("Yes\n");    else        System.out.print("No\n");}} // This code is contributed by shubham

## Python3

 # Python3 implementation of the# above approach  # Function returns true if# s1 is suffix of s2def isSuffix(s1, s2):     n1 = len(s1)    n2 = len(s2)         if(n1 > n2):        return False    for i in range(n1):             if (s1[n1 - i - 1] !=            s2[n2 - i - 1]):            return False;    return True; # Function to check if# binary equivalent of a# number ends in "111" or notdef CheckBinaryEquivalent(N, s):      # To store the binary    # number    B_Number = 0;    cnt = 0;      while (N != 0):          rem = N % 2;        c = pow(10, cnt);        B_Number += rem * c;        N //= 2;          # Count used to store        # exponent value        cnt += 1;         bin = str(B_Number);       return isSuffix(s, bin);   # Driver code   if __name__ == "__main__":         N = 23;    s = "111";         if (CheckBinaryEquivalent(N, s)):        print("Yes")    else:        print("No") # This code is contributed by rutvik_56

## C#

 // C# implementation of the // above approach using System;using System.Collections; class GFG{ // Function returns true if// s1 is suffix of s2static bool isSuffix(String s1, String s2){    int n1 = s1.Length,        n2 = s2.Length;             if (n1 > n2)        return false;             for(int i = 0; i < n1; i++)        if (s1[n1 - i - 1] !=            s2[n2 - i - 1])        return false;             return true;} // Function to check if binary equivalent// of a number ends in "111" or notstatic bool CheckBinaryEquivalent(int N,                                  String str){         // To store the binary    // number    int B_Number = 0;    int cnt = 0;     while (N != 0)    {        int rem = N % 2;        int c = (int)Math.Pow(10, cnt);        B_Number += rem * c;        N /= 2;         // Count used to store        // exponent value        cnt++;    }    String bin = B_Number.ToString();    return isSuffix(str, bin);} // Driver Codepublic static void Main (String[] args){    int N = 23;    String str = "111";         if (CheckBinaryEquivalent(N, str))        Console.WriteLine("Yes\n");    else        Console.WriteLine("No\n");}} // This code is contributed by jana_sayantan

## Javascript


Output:
Yes

Time Complexity: O(|str)

Auxiliary Space: O(1)

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